What is this expression, precisely? : $2^x+3^x+4^x+5^x+6^x+7^x$

Question

$2^x+3^x+4^x+5^x+6^x+7^x=P$

I have trouble understanding what this number $P$ actually is…

I suspect it might be a sum of some sort either arithmetic or geometric, however this nasty placement of an x here makes me doubt my predictions…

So what it is, exactly?

Answer 1

In fact $$\sum_{k=2}^n k^x=H_n^{(-x)}-1$$ where appear generalized harmonic numbers.

If you know $x$, you easily compute $P$. If you know $P$, you can compute $x$ but, in the most general case, this would require a numerical method.

If you are concerned by the second point, let me know and I shall eleborate.

Answer 2

This is neither a geometric sum, nor an arithmetic sum. It is just a sum. You cannot do much, we do not know the value of P, and neither of x. We cannot factor this, either— but you can simplify it, however, not in any useful way.

Answer 3

As others have already pointed out, this is neither an arithmetic nor geometric sum. As far as a closed form for P, I present this approximation:

Let $2^x+3^x+4^x+5^x+6^x+7^x=P$, where P is a function of x

Then, take the derivative of both sides:

$2^xln(2)+3^xln(3)+4^xln(4)+5^xln(5)+6^xln(6)+7^xln(7)=P'$

Now, if we could factor out those natural logs it would be great, but unfortunately we can't, so we can approximate by dividing by their average:

$\frac{2^xln(2)+3^xln(3)+4^xln(4)+5^xln(5)+6^xln(6)+7^xln(7)}{(ln(2)+ln(3)+ln(4)+ln(5)+ln(6)+ln(7))/6} = \frac{P'}{(ln(2)+ln(3)+ln(4)+ln(5)+ln(6)+ln(7))/6}$

Which by the log properties is equal to:

$\frac{2^xln(2)+3^xln(3)+4^xln(4)+5^xln(5)+6^xln(6)+7^xln(7)}{ln(7!)/6} = \frac{P'}{{ln(7!)/6}}$

The left hand side then approximately cancels out the natural logs like so:

$2^x+3^x+4^x+5^x+6^x+7^x \approx \frac{P'}{{ln(7!)/6}}$

Substitute back in P to get this differential equation:

$P \approx \frac{P'}{{ln(7!)/6}}$

(I assume you know to solve a differential equation like this) Solve it to get:

$ln(P)\approx\frac{ln(7!)}{6}x+C$

Then raise e to power of both sides and simplify to find that $P\approx 7!^{\frac x6+C}$

C is a some constant - try twiddling with this equation on Desmos to see how well it approximates.